Question: Let $\mathcal{S}_{n}$ be the set of strings with only 0's or 1's with length $n$ such that any 3 adjacent place numbers sum to at least 1. For example, $00100$ works, but $10001$ does not. Find the number of elements in $\mathcal{S}_{11}$.

Solution: We will solve this problem by constructing a recursion satisfied by $\mathcal{S}_n$.
Let $A_1(n)$ be the number of such strings of length $n$ ending in 1, $A_2(n)$ be the number of such strings of length $n$ ending in a single 0 and $A_3(n)$ be the number of such strings of length $n$ ending in a double zero. Then $A_1(1) = 1, A_2(1) = 1, A_3(1) = 0, A_1(2) = 2, A_2(2) = 1$ and $A_3(2) = 1$.
Note that $\mathcal{S}_n = A_1(n) + A_2(n) + A_3(n)$. For $n \geq 2$ we have $A_1(n) = \mathcal{S}_{n - 1} = A_1(n - 1) + A_2(n - 1) + A_3(n - 1)$ (since we may add a 1 to the end of any valid string of length $n - 1$ to get a valid string of length $n$), $A_2(n) = A_1(n -1)$ (since every valid string ending in 10 can be arrived at by adding a 0 to a string ending in 1) and $A_3(n) = A_2(n - 1)$ (since every valid string ending in 100 can be arrived at by adding a 0 to a string ending in 10).
Thus $\mathcal{S}_n = A_1(n) + A_2(n) + A_3(n) = \mathcal{S}_{n - 1} + A_1(n - 1) + A_2(n - 1) = \mathcal{S}_{n -1} + \mathcal{S}_{n - 2} + A_1(n - 2) = \mathcal{S}_{n - 1} + \mathcal{S}_{n -2} + \mathcal{S}_{n - 3}$. Then using the initial values $\mathcal{S}_1 = 2, \mathcal{S}_2 = 4, \mathcal{S}_3 = 7$ we can easily compute that $\mathcal{S}_{11} = \boxed{927}$.